/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Pollutants in auto exhausts. In ... [FREE SOLUTION] | 91Ó°ÊÓ

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Pollutants in auto exhausts. In 2017, the entire fleet of light-duty vehicles sold in the United States by each manufacturer must emit an average of no more than 86 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) over the useful life ( 150,000 miles of driving) of the vehicle. NOX + NMOG emissions over the useful life for one car model vary Normally with mean \(80 \mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). (a) What is the probability that a single car of this model emits more than 86 \(\mathrm{mg} / \mathrm{mi}\) of NOX + NMOG? (b) A company has 25 cars of this model in its fleet. What is the probability that the average \(\mathrm{NOX}+\mathrm{NMOG}\) level \(\mathrm{x}^{-}=\)of these cars is above \(86 \mathrm{~g} / \mathrm{mi}\) ?

Short Answer

Expert verified
(a) 6.68%; (b) ~0%

Step by step solution

01

Understand the Problem

You need to find two probabilities where emission data follows a normal distribution. In part (a), you determine the likelihood of a single car emitting more than 86 mg/mi. In part (b), you calculate the probability that the average emissions from 25 cars exceed this value.
02

Identify Given Data

From the problem, the emissions mean is \( \mu = 80 \, \text{mg/mi} \) and the standard deviation is \( \sigma = 4 \, \text{mg/mi} \). The emissions threshold is 86 mg/mi. For part (b), a sample size of 25 cars is used.
03

Calculate the Required Z-score for (a)

For a single car, use the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest (86 mg/mi). Thus, the Z-score is \[ Z = \frac{86 - 80}{4} = 1.5. \]
04

Find Probability for (a) Using Z-score

Consult the standard normal distribution table to find the probability of a Z-score greater than 1.5. This value is approximately 0.0668. So, the probability that a single car emits more than 86 mg/mi is about 6.68%.
05

Calculate Standard Error for (b)

For the average emissions from 25 cars, the standard error is \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( n = 25 \). Thus, \[ \sigma_{\bar{x}} = \frac{4}{\sqrt{25}} = 0.8. \]
06

Calculate the Required Z-score for (b)

Use the same formula, but adjust for the standard error: \( Z = \frac{X - \mu}{\sigma_{\bar{x}}} \). Thus, \[ Z = \frac{86 - 80}{0.8} = 7.5. \]
07

Find Probability for (b) Using Z-score

A Z-score of 7.5 is extremely high, indicating that the probability of the average emissions from 25 cars being above 86 mg/mi is practically 0. Emission exceeding this level is nearly impossible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is the measure of the likelihood that a particular event will occur. In the context of the emissions problem, we are interested in two probabilities:
- The probability that a single car emits more than 86 mg/mi
- The probability that the average emission level of 25 cars exceeds 86 mg/mi.
When we talk about probabilities in terms of a normal distribution, we refer to outcomes within the context of a bell curve, where certain values have specific probabilities of occurrence. This is crucial in determining how likely an emission level is to exceed specific limits. For our single car model, the computation involves locating where 86 mg/mi falls under this curve.
Similarly, when taking a fleet of 25 cars into account, the scenario shifts to evaluating how often their collective average would surpass the threshold, giving us insights into broader impacts.
Z-score
A Z-score indicates how many standard deviations an element is from the mean of a distribution. It is calculated as follows:
\[ Z = \frac{X - \mu}{\sigma} \]
For individual (a) car emissions, X is 86 mg/mi, \(\mu\) is 80 mg/mi, and \( \sigma \) is 4 mg/mi. Calculating gives a Z-score of 1.5, which tells us that 86 mg/mi is 1.5 standard deviations above the average emission rate.
For (b) the fleet average emissions, we adjust the formula to use the standard error, if the sample size \(n\) is applied. In this case, a Z-score of 7.5 indicates that such a high average is extraordinarily unlikely under normal circumstances. All Z-scores provide a way to traverse probability tables and ascertain the likelihood of specific outcomes.
Standard Deviation
Standard deviation (SD) is a measure of variability or dispersion in a dataset. In simpler terms, it tells us how spread out numbers are around their average. A smaller standard deviation means data points are closer to the mean, while a larger one indicates they are spread out over a wider range.
In the exercise, the standard deviation of the emissions is given as 4 mg/mi. This helps us understand how consistent or variable the emission levels are from car to car. When calculating the Z-score, the standard deviation plays a crucial role in determining how unusual a data point (like an emission level) is in relation to the mean.
For groups, like the 25 cars, we use the standard deviation to compute the standard error, which gives us a precise understanding of the sample mean's variability, providing further insights into probability assessments.
Sample Size
Sample size is the number of observations or data points collected in a sample. The size of the sample significantly impacts the reliability of statistical measures, like the average, due to the law of large numbers, which suggests that larger samples produce more reliable estimates of the population mean.
In context (a) with just one car, we focus on individual emissions. But in context (b) when we consider 25 cars, we enter the realm of sampling distributions. Here, the sample size directly influences the calculation of the standard error, represented as:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
A larger sample size results in a smaller standard error, leading to more precise estimates of the population mean. Thus, the fleet's average emissions offer a different probability landscape due to the stabilizing effect of a larger (n) number.

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Most popular questions from this chapter

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